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Find the exact value of the expression.
Exercise (a)
$\sin^{-1}(\frac{-\sqrt{3}}{2})$
Part 1 of 2
If $y = \sin^{-1}x$, then $\sin y = x$, $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Therefore, to find $y = \sin^{-1}(\frac{-\sqrt{3}}{2})$, we must find an angle $y$ whose sine is $-\frac{\sqrt{3}}{2}$.
There are many possible angles with this sine, but the range of $y = \sin^{-1}x$ is restricted to $[\frac{-\pi}{2}, \frac{\pi}{2}]$, and so $y$ must be in this interval.
Part 2 of 2
The angle $y$ whose sine is $-\frac{\sqrt{3}}{2}$ and which is in the interval $[\frac{-\pi}{2}, \frac{\pi}{2}]$ is $y = $ radians.
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